Chapter 8 Mechanical Failurehfang2/MECE3345/Lectures/Chapter8.pdf · DESIGN AGAINST CRACK GROWTH...

Materials Science Chapter 8 1

Chapter 8 Mechanical Failure

• How do flaws in a material initiate failure?• How is fracture resistance quantified; how do different

material classes compare?• How do we estimate the stress to fracture?• How do loading rate, loading history, and temperature

affect the failure stress?

Ship-cyclic loadingfrom waves.

Computer chip-cyclicthermal loading.

Hip implant-cyclicloading from walking.


DUCTILE VS BRITTLE FAILURE

Very Ductile

Moderately Ductile BrittleFracture

behavior:

Large Moderate%AR or %EL: Small• Ductile

fracture isdesirable!

• Classification:

Ductile:warning before

fracture

Brittle:No

warning


EX: FAILURE OF A PIPE

• Ductile failure:--one piece--large deformation

• Brittle failure:--many pieces--small deformation


MODERATELY DUCTILE FAILURE• Evolution to failure:

necking void nucleation

void growth and linkage

shearing at surface

fracture

σ

• Resultingfracturesurfaces(steel)

50 µm

Particles serve as void nucleationsites.

50 µm

100 µm

Fracture surface of tire cord wire loaded in tension.


BRITTLE FRACTURE SURFACES• Intergranular(between grains)

• Intragranular(within grains)

316 S. Steel (metal)

304 S. Steel (metal)

4 mm 160µm

Al Oxide(ceramic)Polypropylene

(polymer)

3µm1 mm


IDEAL VS REAL MATERIALS

• Stress-strain behavior (Room T):

σ

ε

E/10

E/100

0.1

perfect mat’l-no flaws

carefully produced glass fiber

typical ceramic typical strengthened metaltypical polymer

TS << TSengineeringmaterials

perfectmaterials

• DaVinci (500 yrs ago!) observed...--the longer the wire, the

smaller the load to fail it.• Reasons:

--flaws cause premature failure.--Larger samples are more flawed!


FLAWS ARE STRESS CONCENTRATORS!• Elliptical hole in

a plate:• Stress distrib. in front of a hole:

• Stress conc. factor:

BAD! Kt>>3NOT SO BAD

Kt=3

σmax

ρt

≈ 2σo

a + 1⎛ ⎝ ⎜

⎞ ⎠ ⎟

ρtσ

σo

2a

Kt = σmax /σo

• Large Kt promotes failure:


ENGINEERING FRACTURE DESIGN• Avoid sharp corners!

r/h

sharper fillet radius

increasing w/h

0 0.5 1.01.0

1.5

2.0

2.5

Stress Conc. Factor, Ktσ

maxσ

o=

r , fillet

radius

w

h

σo

σmax


• ρt at a cracktip is verysmall!

σ

• Result: crack tipstress is very large.

• Crack propagates when:the tip stress is largeenough to make: distance, x,

from crack tip

σtip = K2π x

σtip

increasing K

K ≥ Kc

WHEN DOES A CRACK PROPAGATE?


FRACTURE TOUGHNESS•Griffith theory of brittle fracture

)2(a

E sc π

γσ =

Modulus of elasticity

Specific surface energy

Critical stress for crack propagationOne half the length of an internal crack

•Stress intensity factor

aYK πσ=Dimensionless function Y (crack and specimen size and geometries, load…)•Fracture toughness

aYK cc πσ=


GEOMETRY, LOAD, & MATERIAL

• Condition for crack propagation:

• Values of K for some standard loads & geometries:σ

2a2a

σ

aa

K = σ πa K = 1.1σ πa

K ≥ KcStress Intensity Factor:--Depends on load &

geometry.

Fracture Toughness:--Depends on the material,

temperature, environment, &rate of loading.

units of K :MPa mor ksi in


DESIGN AGAINST CRACK GROWTH• Crack growth condition:

Yσ πa

K ≥ Kc

• Largest, most stressed cracks grow first!

--Result 2: Design stressdictates max. flaw size.

--Result 1: Max flaw sizedictates design stress.

amax <

1π

KcYσdesign

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

σdesign <

KcY πamax

amax

σno fracture

fracture

amax

σ

no fracture

fracture


DESIGN EX: AIRCRAFT WING

• Two designs to consider...Design A--largest flaw is 9 mm--failure stress = 112 MPa

Design B--use same material--largest flaw is 4 mm--failure stress = ?

• Use... σc =

KcY πamax

• Key point: Y and Kc are the same in both designs.--Result:

σc amax( )A = σc amax( )B

9 mm112 MPa 4 mm

Answer: σc( )B = 168MPa• Reducing flaw size pays off!

• Material has Kc = 26 MPa-m0.5


• Increased loading rate...--increases σy and TS--decreases %EL

• Why? An increased rategives less time for disl. tomove past obstacles.

initial heightfinal height

sample

σ

ε

σy

σy

TS

TSlarger ε

smaller ε

(Charpy)• Impact loading:--severe testing case--more brittle--smaller toughness

LOADING RATE


TEMPERATURE• Increasing temperature...

--increases %EL and Kc

• Ductile-to-brittle transition temperature (DBTT)...

BCC metals (e.g., iron at T < 914C)

Imp

ac

t E

ne

rgy

Temperature

FCC metals (e.g., Cu, Ni)

High strength materials (σy>E/150)

polymers

More Ductile Brittle

Ductile-to-brittle transition temperature


DESIGN STRATEGY:STAY ABOVE THE DBTT!

• Pre-WWII: The Titanic • WWII: Liberty ships

Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(b), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Earl R. Parker, "Behavior of Engineering Structures", Nat. Acad. Sci., Nat. Res. Council, John Wiley and Sons, Inc., NY, 1957.)

Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(a), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Dr. Robert D. Ballard, The Discovery of the Titanic.)

• Problem: Used a type of steel with a DBTT ~ Room temp.


• Fatigue = failure under cyclic stress.

tension on bottom

compression on top

countermotor

flex coupling

bearing bearing

specimen

• Stress varies with time.

• Key points: Fatigue...--can cause part failure, even though σmax < σc.--causes ~ 90% of mechanical engineering failures.

FATIGUE


FATIGUE KEY PARAMETERS•Mean stress

•Range of stress2

minmax σσσ +=m

minmax σσσ −=r

•Stress amplitude

•Stress ratio22

minmax σσσσ −== r

a

max

min

σσ

=R


The S-N CURVE

aS σ=

Fatigue strength, fatigue life, and fatigue limit


FATIGUE DESIGN PARAMETERS

• Fatigue limit, Sfat:--no fatigue if S < Sfat

• Sometimes, thefatigue limit is zero!

Sfat

case for steel (typ.)

N = Cycles to failure103 105 107 109

unsafe

safe

S = stress amplitude

case for Al (typ.)

N = Cycles to failure103 105 107 109

unsafe

safe



FATIGUE MECHANISM•Crack initiation

•Crack propagation


FATIGUE MECHANISM• Crack grows incrementally

dadN

= ∆K( )mtypical 1 to 6

( ) ))((~ minmax aYKa πσσσ −=∆∆increase in crack length per loading cycle

• Failed rotating shaft--crack grew even though

Kmax < Kc--crack grows faster if

• ∆σ increases• crack gets longer• loading frequency

increases.

crack origin


IMPROVING FATIGUE LIFE

1. Impose a compressivesurface stress(to suppress surfacecracks from growing)

N = Cycles to failure

moderate tensile σmlarger tensile σm


near zero or compressive σm

--Method 1: shot peening

2. Remove stressconcentrators.

bad

bad

better

better

--Method 2: carburizing

C-rich gasput

surface into

compression

shot


CREEP

• Occurs at elevated temperature, T > 0.4 Tmelt• Deformation changes with time.

σ,εσ

0 t


CREEP

•Temperature dependence•Stress dependence

timeelastic

primary secondary

tertiary

T < 0.4 Tm

INCREASING T

0

strain, ε


SECONDARY CREEP• Strain rate is constant at a given T, σ

--strain hardening is balanced by recovery

stress exponent (material parameter)

strain rateactivation energy for creep(material parameter)

applied stressmaterial const. εs = K2σn exp −

QcRT

⎛

⎝ ⎜

⎞

⎠ ⎟

.

10

20

40

100

200

Steady state creep rate (%/1000hr)10-2 10-1 1

ε s

Stress (MPa)427C

538C

649C

• Strain rateincreasesfor larger T, σ


CREEP FAILURE• Failure:

along grain boundaries.

time to failure (rupture)

function ofapplied stress

temperature

T(20 + log t r ) = L

L(103K-log hr)

Str

ess

, ksi

100

10

112 20 24 2816

data for S-590 Iron

20appliedstress

g.b. cavities

• Time to rupture, tr

• Estimate rupture timeS 590 Iron, T = 800C, σ = 20 ksi

T(20 + log t r ) = L

1073K

24x103 K-log hr

Ans: tr = 233hr

Chapter 8 Mechanical Failurehfang2/MECE3345/Lectures/Chapter8.pdf · DESIGN AGAINST CRACK GROWTH...

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